0
$\begingroup$

The game works like this,

Suppose, there is a total of 900 numbers(1.00 to 9.99) from you can choose any one number to bet.

Example:

1) If you choose 1.10 number and place a bet of 100$ on it. If the generated random number is 1.10 or greater than that then you will receive 110 in return(100 * 1.10).

2) If you place a 100$ bet on number 5.00 and again generated random number is 5.00 or greater then that then you will receive 500 (100 * 5.00) in reward.

But the twist here is there is less probability for a higher number to generate a then lower number,

1) The probability of the generated number is between 1.00 - 1.49 is 60%

2) For numbers between 1.50 - 1.99. There is a probability of 30%.

3) For rest 2.00 - 9.99, there is a probability of 10%.

You can refer to this link for getting a better understanding of the game: Moon-Game

My question is if I bet on any particular number then how can I calculate my win probability?

Your answers appreciated. Thanks in advance.

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Are all numbers within your brackets equally likely to be chosen? $\endgroup$ – Stephan Kolassa Jul 25 '19 at 12:52
  • $\begingroup$ yes, if the chosen number is 9.99 then reward will be bet * 9.99 incase you win $\endgroup$ – Meet Siraja Jul 25 '19 at 12:53
  • 1
    $\begingroup$ That is not what I asked. My question is: are 1.11 and 1.34 equally likely (namely, with a chance of $0.60\times \frac{1}{50}$? $\endgroup$ – Stephan Kolassa Jul 25 '19 at 12:54
  • 1
    $\begingroup$ Why is the information about payoffs relevant when the question is only about probabilities? Are you sure you have formulated the question you intended to ask? $\endgroup$ – whuber♦ Jul 25 '19 at 12:55
  • $\begingroup$ Yes stephan, both numbers are equally likely to be chosen. If payout info is not relevant you can ignore it whuber. I just thought that it will be helpful for better understanding. $\endgroup$ – Meet Siraja Jul 25 '19 at 13:07
1
$\begingroup$

I'm taking this question to mean that if you pick a number at random, what is the probability you'll win?

This is simply a weighted average. There's a 5/90 chance the number is chosen from bracket 1, 5/90 for bracket 2, and 8/9 for the last bracket. Multiply each case by it's winning probability and you have your answer. So,

(5/90)X(6/10) + (5/90)X(3/10) + (8/9)X(1/10) = 0.1389.

This is how the game creators make their money, as with enough people betting, only around 14% of them will win. The rest of the money goes to them.

Hope this helps!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did you calculate the chance for each bracket like 5/90, 5/90 and 8/9 $\endgroup$ – Meet Siraja Jul 25 '19 at 13:31
  • $\begingroup$ 50 of the 900 numbers are between 1.00 and 1.49. 50/900 simplifies to 5/90. Likewise for 1.50 to 1.99. The remaining 800 out of 900 numbers is 800/900 which simplifies to 8/9. $\endgroup$ – GridAlien Jul 25 '19 at 13:33
  • $\begingroup$ Okay, I get you. $\endgroup$ – Meet Siraja Jul 25 '19 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.